Evaluate sum(n=2,n=infinity) ln(1-1/n^2) if it exists.

Answer 1

1-1/n^2 = (n^2 - 1)/n^2 =(n+1)(n-1)/n^2

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Log(1-1/n^2) =

Log(n+1) + Log(n-1) - 2 Log(n).

If we define the difference operator Delta as:

(Delta f)(n) = f(n+1) - f(n)

Then applying Delta twice gives:

(Delta^2 f)(n) =

f(n+2) - 2f(n+1) + f(n)

The formula for Log(1-1/n^2) has the same structure. We see that:

Log(1-1/n^2) = Delta^2 f(n)

with:

f(n) = log(n-1)

In general, we have:

Sum from n = r to s of (Delta g)(n) =

g(r+1) - g(r) + g(r+2) - g(r+1) + ...
+ g(s) - g(s-1) + g(s+1) - g(s) =

g(s+1) - g(r)

So, we have:

Sum from n = 2 to N of Log(1-1/n^2) =

Sum from n = 2 to N of
Delta(Delta f)(n) =

(Delta f)(N+1) - (Delta f)(2)

Now

f(n) = log(n-1),

so

(Delta f)(n) = Log(n) - log(n-1) =

Log[n/(n-1)]

and we see that:

(Delta f)(N+1) - (Delta f)(2) =

Log[(N+1)/N] - Log(2)

The sum to infinity is the limit for N to infinity. We can write:

Log[(N+1)/N] = Log(1+1/N) =
1/N + O(1/N^2)

So, this terms goes to zero in the limit N to infinity.

The summation is thus equal to -Log(2)